Do falling objects drop at the same rate (for instance a pen and a bowling
ball dropped from the same height) or do they drop at different rates? I know a feather floats down very slowly but I would think a heavy object would fall faster than
a light object. Thanks for your help. I have a bet on this one.

Asked by: Terri

Answer

If no air resistance is present, the rate of descent depends only on
how far the object has fallen, no matter how heavy the object is. This means that two objects will reach the ground at the same time if they are dropped
simultaneously from the same height. This statement follows from the law of
conservation of energy and has been demonstrated experimentally by dropping
a feather and a lead ball in an airless tube.

When air resistance plays a role, the shape of the object becomes
important. In air, a feather and a ball do not fall at the same rate. In the
case of a pen and a bowling ball air resistance is small compared to the
force a gravity that pulls them to the ground. Therefore, if you drop a pen and a bowling ball you could probably not tell
which of the two reached the ground first unless you dropped them from a
very very high tower.
Answered by: Dr. Michael Ewart, Researcher at the University of Southern California

The above answer is perfectly correct, but, this is a question that
confuses many people, and they are hardly satisfied by us
self-assured physcists' answers. There is one good explanation
which makes everybody content -- which does not belong to me, but
to some famous scientist but I can't remember whom (Galileo?);
and I think it would be good to have it up here.

(The argument has nothing to do with air resistance, it is
assumed to be absent. The answer by Dr. Michael Ewart answers
that part already.)

The argument goes as follows: Assume we have a 10kg ball and a
1kg ball. Let us assume the 10kg ball falls faster than the 1kg
ball, since it is heavier. Now, lets tie the two balls together.
What will happen then? Will the combined object fall slower,
since the 1kg ball will hold back the 10kg ball? Or will the
combination fall faster, since it is now an 11kg object? Since
both can't happen, the only possibility is that they were falling
at the same rate in the first place.

Sounds extremely convincing. But, I think there is a slight
fallacy in the argument. It mentions nothing about the nature of
the force involved, so it looks like it should work with any kind
of force! However, it is not quite true. If we lived on a world
where the 'falling' was due to electrical forces, and
objects had masses and permanent charges, things would be
different. Things with zero charge would not fall no matter what
their mass is. In fact, the falling rate would be proportional to
q/m, where q is the charge and m is the mass. When you tie two
objects, 1 and 2, with charges q_{1}, q_{2}, and m_{1}, m_{2}, the combined
object will fall at a rate (q_{1}+q_{2})/(m_{1}+m_{2}). Assuming
q_{1}/m_{1} < q_{2}/m_{2}, or object 2 falls faster than object one, the
combined object will fall at an intermediate rate (this can be
shown easily). But, there is another point. The 'weight' of an
object is the force acting on it. That is just proportional to q,
the charge. Since what matters for the falling rate is q/m, the
weight will have no definite relation to rate of fall. In fact,
you could have a zero-mass object with charge q, which will fall
infinitely fast, or an infinite-mass object with charge q, which
will not fall at all, but they will 'weigh' the same! So, in
fact, the original argument should be reduced to the following
statement, which is more accurate:

If all objects which have equal weight fall at the same rate,
then _all_ objects will fall at the same rate, regardless of
their weight.

In mathematical terms, this is equivalent to saying that if
q_{1}=q_{2} then m_{1}=m_{2} or, q/m is the same for all objects, they will
all fall at the same rate! All in all, this is pretty hollow an
argument.

Going back to the case of gravity.. The gravitational force is

( G is a constant, called constant of gravitation, M is the mass
of the attracting body (here, earth), and m_{1} is the
'gravitational mass' of the object.)

And newton's law of motion is

where m_{2} is the 'inertial mass' of the object, and a is the
acceleration.

Now, solving for acceleration, we find:

Which is proportional to m_{1}/m_{2}, i.e. the gravitational mass
divided by the inertial mass. This is our old 'q/m' from the
electrical case! Now, if and only if m_{1}/m_{2} is a constant for all
objects, (this constant can be absorbed into G, so the question
can be reduced to m_{1}=m_{2} for all objects) they will all fall at
the same rate. If this ratio varies, then we will have no
definite relation between rate of fall, and weight.

So, all in all, we are back to square one. Which is just
canceling the masses in the equations, thus showing that they
must fall at the same rate. The equality of the two masses is a
necessity for general relativity, and enters it naturally. Also,
the two masses have been found to be equal to extremely good
precision experimentally. The correct answer to the question 'why
objects with different masses fall at the same rate?' is,
'beacuse the gravitational and inertial masses are equal for all
objects.'

Then, why does the argument sound so convincing? Since our daily
experience and intuition dictates that things which weigh the
same, fall at the same rate. Once we assume that, we have
implicitly already assumed that the gravitational mass is equal
to the inertial mass. (Wow, what things we do without noticing!).
The rest of the argument follows easily and naturally...
Answered by: Yasar Safkan, Physics Ph.D. Candidate, M.I.T.

'A theory with mathematical beauty is more likely to be correct than an ugly one that fits some experimental data. God is a mathematician of a very high order, and He used very advanced mathematics in constructing the universe.'